Question

我有一个示例数据帧df和一个数组n，如下所示。我想根据索引中的数组值进行过滤。输出数据帧也如下所示。我尝试了Out = df[df.index == n]，Out = df.loc[df.index == n]和df.loc[n]，但均无法正常工作，并给出错误Lengths must match to compare。谁能帮我解决这个问题。这里的数组是对应于数据帧的行号。

df = 
             Open   High    Low    Close    Adj Close   Volume
2007-06-18  0.33979 0.33979 0.33979 0.33979 0.33979 1591888
2007-06-29  0.33074 0.33074 0.33074 0.33074 0.33074 88440
2007-06-20  0.33526 0.33526 0.33526 0.33526 0.33526 3538
2007-06-21  0.32113 0.32113 0.32113 0.32113 0.32113 3550
2007-06-22  0.34713 0.34713 0.34713 0.34713 0.34713 670
2007-06-16  0.33979 0.33979 0.33979 0.33979 0.33979 1591888
2007-06-30  0.33074 0.33074 0.33074 0.33074 0.33074 88440
2007-06-31  0.33526 0.33526 0.33526 0.33526 0.33526 3538
2007-06-44  0.32113 0.32113 0.32113 0.32113 0.32113 3550
2007-06-22  0.34713 0.34713 0.34713 0.34713 0.34713 670

n = array([0, 1, 2, 3])

Out  = 
            Open      High  Low     Close   Adj Close   Volume
2007-06-18  0.33979 0.33979 0.33979 0.33979 0.33979 1591888
2007-06-29  0.33074 0.33074 0.33074 0.33074 0.33074 88440
2007-06-20  0.33526 0.33526 0.33526 0.33526 0.33526 3538
2007-06-21  0.32113 0.32113 0.32113 0.32113 0.32113 3550

Answer 1

用于切片的熊猫符号：

df.iloc[0:4,:]

Answer 2

使用DataFrame.iloc用于按位置选择：

.subscribe((Consumer<? super List<String>>)getAnimalsObserver());

Answer 3

用输入内容替换<>之间的所有内容

# slice by column position
df.iloc[<start_row>:<end_row>, <column_start_position>:<column_end_position>]
# for everything in a column
df.iloc[:, <column_position>]


# slice by column name
df.loc[<start_row>:<end_row>, <column_name>]
# for everything in a column
df.loc[:, <column_name>]

也请在pandas文档中查看Index and Selecting Data。超级丰富，即使在第一次通过时也不会造成混淆。

在python中通过其索引和数组切片数据帧

3 个答案: